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Math Help - Boolean algebra help

  1. #1
    Member Jones's Avatar
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    Boolean algebra help

    Hi,

    I have a boolean function that im not sure how to solve.

    (x+z')'*(y+z')' + xy' +xz'

    im supposed to write this on disjunctive normal form.

    I tried multiplying the xy' and the xz' term
    by (z+z') and (y+y') but that makes the whole expression really long.

    Any ideas?

    Jones
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  2. #2
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    A start to rewriting this

    I assume (x+z')' means not ( x or (not z)). If so, then doesn't this simplify to
    x'z? Then your expression simplifies to:

    = x'zy'z + xy' + xz'
    = x'y'z + xy' + xz'
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  3. #3
    Member Jones's Avatar
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    No, thats incorrect.

    The correct answer should be x'y'z + xy'z' + xy'z + xyz'
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  4. #4
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    A little more manipulation

    Note:
    xy' = xy'z + xy'z'
    because z + z'=1
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  5. #5
    Member Jones's Avatar
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    So,


    Simplifying the first two paranthesis
    \overline{x+\overline{z}} \ast \overline{y+\overline{z}}

    Should simplify down to just \overline{x}z \ast \overline{y}z

    Applying Morgans law gives:

    \overline{\overline{x}z \ast \overline{y}z} \rightarrow x+\overline{z} + y+\overline{z}

    Using de morgans again gives: \overline{x}z+\overline{y}z

    The whole expression now becomes \overline{x}z+\overline{y}z+x\overline{y}+x\overli  ne{z}

    Is this correct?
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  6. #6
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    you need my advice it is a 3 literals function so it is easy to build the truth table and see where the function set to 1 then build the karnough map and you will get the most simplified form
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  7. #7
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    the simplified form is z'x+zy'

    waiting your reply

    truth table :
    X Y Z F
    0 0 0 0
    0 0 1 1
    0 1 0 0
    0 1 1 0
    1 0 0 1
    1 0 1 1
    1 1 0 1
    1 1 1 0

    K-map :
    xy 00 01 11 10
    z 0 0 0 (1 1)
    1 1) 0 0 (1
    => z'x+zy'
    Last edited by Jameson; November 24th 2009 at 06:55 AM.
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